Question

The atomic number of an element $M$ is \[26\]. How many electrons are present in the \[M - shell\] of the element in its \[{M^{3 + }}\] state?
A. \[11\]
B. \[15\]
C. \[14\]
D. \[13\]

Answer 1

Hint:
The atomic number \[26\] in the periodic table is for the element \[Fe\] known as Iron, and \[K,{\text{ }}L,M,N\]are the shells in which the electron are filled for which the value is as follows:
\[K = 1,{\text{ }}L = 2,{\text{ }}M = 3,{\text{ }}N = 4\]and so on.

Complete answer:
For this question one must know what are the rules for filling the electrons to obtain electronic configurations;
i. Aufbau Principle: The electron which is added will always occupy the orbital with the lowest energy first.
ii. Pauli Exclusion Principle: Each orbital can hold a maximum of two electrons of opposite spins.
iii. Hund’s Rule of Multiplicity: When filling a sub-shell, each orbital must be occupied singly (keeping electron spins the same) before they are occupied in pairs.
And the electrons are filled in the order as follows: \[1s < 2s < 2p < 3s < 3p < 4s < 3d\] and further. The \[1,{\text{ }}2\] and \[3\] in the configuration are the \[K,{\text{ }}L,{\text{ }}M\] shells and the \[s,{\text{ }}p,{\text{ }}d\] are the sub-shells where the \[s\] sub-shell can have \[2\] electrons, \[p\] sub-shell can have \[6\] electrons and \[d\] sub-shell can hold a maximum of \[10\] electrons.
Now for the element \[26\] the electronic configuration will be as follows:
Electronic configuration is \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^6}\]
Now as we are asked for the \[{M^{3 + }}\] state which means we have to remove three electrons and the first \[2\] electrons are removed from the \[4s\] then from the 3d this is because as we know that the \[4s\] has lower energy than \[3d\] but still \[4s\] is the outermost electron as its experiences repulsion, so the \[{M^{3 + }}\] electronic configuration will be
\[{M^{3 + }}\] electronic configuration will be \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^5}\]
Now having a look at the configuration the electrons present for\[M{\text{ }} = {\text{ }}3\] , the number of electrons present will be \[2 + 6 + 5 = 13\;\] electrons.

Hence, the correct option is D.

Note:The energy for \[4s\] is less than \[3d\] because the overall \[\left( {n + l} \right)\] value is less for \[4s\] than the \[3d\], now the value of \[l\] means the value of sub-shells; \[s,{\text{ }}p,{\text{ }}d\]and \[f\]are \[0,{\text{ }}1,{\text{ }}2,{\text{ }}3\] respectively and for \[4s\] the \[\left( {n + l} \right)\] value comes out to be \[4 + 0{\text{ }} = {\text{ }}4\] and for \[3d\] is \[3 + 2{\text{ }} = {\text{ }}5.\]